Problem: Divide the following complex numbers. $\dfrac{-2-6i}{-1-3i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-1+3i}$. $ \dfrac{-2-6i}{-1-3i} = \dfrac{-2-6i}{-1-3i} \cdot \dfrac{{-1+3i}}{{-1+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-2-6i) \cdot (-1+3i)} {(-1)^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-2-6i) \cdot (-1+3i)} {(-1)^2 - (-3i)^2} $ $ = \dfrac{(-2-6i) \cdot (-1+3i)} {1 + 9} $ $ = \dfrac{(-2-6i) \cdot (-1+3i)} {10} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-2-6i}) \cdot ({-1+3i})} {10} $ $ = \dfrac{{-2} \cdot {(-1)} + {-6} \cdot {(-1) i} + {-2} \cdot {3 i} + {-6} \cdot {3 i^2}} {10} $ $ = \dfrac{2 + 6i - 6i - 18 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{2 + 6i - 6i + 18} {10} = \dfrac{20 + 0i} {10} = 2 $